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Answer :
Mean difference (d¯) = 0.32 and standard deviation (sd) ≈ 0.3036.
To find the mean difference (d¯) and standard deviation (sd) of the paired data.
Calculate the differences between each pair of temperatures (12 AM - 8 AM).
Find the mean and standard deviation of these differences.
The differences for each pair:
1. 99.2 - 98.5 = 0.7
2. 99.8 - 99.2 = 0.6
3. 97.7 - 97.4 = 0.3
4. 97.8 - 97.9 = -0.1
5. 97.5 - 97.4 = 0.1
Mean difference (d¯) = (0.7 + 0.6 + 0.3 - 0.1 + 0.1) / 5
Mean difference (d¯) = 1.6 / 5
Mean difference (d¯) = 0.32
Calculate the squared differences from the mean difference:
(0.7 - 0.32)² = 0.1449
(0.6 - 0.32)² = 0.0784
(0.3 - 0.32)² = 0.0009
(-0.1 - 0.32)² = 0.1849
(0.1 - 0.32)² = 0.0516
Find the average of these squared differences:
Average squared difference = (0.1449 + 0.0784 + 0.0009 + 0.1849 + 0.0516) / 5 = 0.09214
Take the square root of the average squared difference to find the standard deviation (sd):
sd = √0.09214
sd ≈ 0.3036
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